(i) To prove the identity \(\cos 4\theta - 4\cos 2\theta \equiv 8\sin^4\theta - 3\):

Use double angle formulas:

\(\cos 4\theta = 2\cos^2 2\theta - 1\)

\(\cos 2\theta = 2\cos^2 \theta - 1\)

Substitute \(\cos 2\theta = 1 - 2\sin^2 \theta\):

\(\cos 4\theta = 2(1 - 2\sin^2 \theta)^2 - 1\)

\(= 2(1 - 4\sin^2 \theta + 4\sin^4 \theta) - 1\)

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1\)

\(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4(1 - 2\sin^2 \theta)\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4 + 8\sin^2 \theta\)

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

(ii) To solve \(\cos 4\theta = 4\cos 2\theta + 3\):

Using the identity \(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 3\), set \(8\sin^4 \theta - 3 = 3\).

\(8\sin^4 \theta = 6\)

\(\sin^4 \theta = \frac{3}{4}\)

\(\sin^2 \theta = \frac{\sqrt{3}}{2}\)

\(\sin \theta = \pm \frac{\sqrt{3}}{2}\)

\(\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ\)

Adjust for the equation \(\cos 4\theta = 4\cos 2\theta + 3\):

\(\theta = 68.5^\circ, 111.5^\circ, 248.5^\circ, 291.5^\circ\)